The graph of $y=ax^2+bx+c$ is given below, where $a$, $b$, and $c$ are integers.  Find $a$.

[asy]
size(140);
Label f;

f.p=fontsize(4);

xaxis(-3,3,Ticks(f, 1.0));

yaxis(-4,4,Ticks(f, 1.0));

real f(real x)

{

return -2x^2+4x+1;

}

draw(graph(f,-.7,2.7),linewidth(1),Arrows(6));
[/asy]
Solution: The vertex of the parabola appears to be at the value $x=1$, where $y=3$.  Therefore we should have \[y=a(x-1)^2+3\] for some integer $a$.  We also know that $(0,1)$ is on the graph of the equation, so \[1=a(0-1)^2+3=a+3.\] Therefore  \[a=1-3=\boxed{-2}.\]